A stone projected vertically upwards with a velocity of 20m/s. Two seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10m/s. Neglecting air resistance, calculate the (i) lenght of time the second stone is in motion before they meet. (ii) velocity of the first stone when they meet (take g=10m/s^2).
1st stone: V = Vo + g*Tr.
0 = 20 - 10Tr, Tr = 2 s. = Rise time.
1. 2nd stone: V = Vo + g*t.
10 = 20 - 10t, t = 1 s. in motion.
2. 1st stone: V = Vo + g*t. = 0 + 10*1 = 10 m/s. When they meet.
To solve this problem, we can use the kinematic equations of motion. Let's start by calculating the time it takes for the two stones to meet.
(i) Length of time the second stone is in motion before they meet:
Let's assume the time it takes for the two stones to meet is 't' seconds after the second stone is projected.
For the first stone:
Initial velocity (u₁) = 20 m/s (upwards)
Acceleration (a) = -10 m/s² (negative because it's acting in the opposite direction to the initial velocity)
Time (t₁) = t + 2 seconds (since the first stone is projected 2 seconds earlier)
Using the equation of motion: v = u + at
Final velocity (v₁) for the first stone when they meet = 0 (because it stops at the highest point)
0 = 20 - 10(t + 2)
10t + 20 = 20
10t = 0
t = 0
So, the second stone is in motion for 0 seconds before they meet.
(ii) Velocity of the first stone when they meet:
To calculate the velocity of the first stone when they meet, we need to find the time it takes for the second stone to reach its velocity of 10 m/s.
For the second stone:
Initial velocity (u₂) = 20 m/s (upwards)
Acceleration (a) = -10 m/s²
Final velocity (v₂) = 10 m/s
Time (t₂) = t (since both stones meet at the same time)
Using the equation of motion: v = u + at
10 = 20 - 10t
10t = 20 - 10
10t = 10
t = 1
So, the first stone meets the second stone after 1 second.
To find the velocity of the first stone when they meet, substitute t = 1 into the equation:
v₁ = u₁ + at
v₁ = 20 - 10(1)
v₁ = 20 - 10
v₁ = 10 m/s
Therefore, the velocity of the first stone when they meet is 10 m/s.
To solve this problem, let's break it down into two parts:
(i) Length of time the second stone is in motion before they meet:
We know that the first stone has a velocity of 20 m/s when it is initially projected upwards. Two seconds later, the second stone is projected with the same initial velocity of 20 m/s. Both stones will eventually meet when they are at their highest point (since they are both projected upwards).
Let's calculate the time it takes for the first stone to reach its highest point:
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration (which in this case is -10 m/s^2 due to gravity), and t is the time, we have:
0 = 20 - (10)t1
10t1 = 20
t1 = 2 seconds
Now, let's calculate the time it takes for the second stone to reach its highest point:
Since the second stone is projected 2 seconds after the first stone, it will take the same amount of time for it to reach its highest point. So t2 = 2 seconds.
Therefore, the length of time the second stone is in motion before they meet is 2 seconds.
(ii) Velocity of the first stone when they meet:
We know that the second stone is rising at a velocity of 10 m/s when they meet. To find the velocity of the first stone when they meet, we need to determine how much time has passed since it was initially projected.
Since the first stone takes 2 seconds to reach its highest point and t1 = 2 seconds, it means that the first stone has already started coming down when they meet. Therefore, the velocity of the first stone when they meet will be its final velocity as it falls from its highest point.
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (which is -10 m/s^2), and t is the time, we can calculate the final velocity as follows:
v = 0 + (10)t1
v = (10)(2)
v = 20 m/s
Therefore, the velocity of the first stone when they meet is 20 m/s.